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Let $y_1, ... y_T$ be a random sample of T = 100 observations on iid Bernoulli distributed random variables $Y_t$ that represent individual decision making where $y_t$ = 1 with probability θ and $y_t$ = 0 with probability (1 - θ).

Τhe density is given by $f(y_t ; θ) = θ^{y_t}(1 - θ)^{1 - y_t}$.

The MLE estimator of of theta is $\hat{θ}$ = $\frac{\sum y_t}{T}$ = $\bar{y}$.

The variance of the MLE estimator is $\hat{σ^2}$ = $\frac{θ(1 - θ)}{Τ}$

I want to use the asymptotic normal approximation to derive an asymptotically valid interval for $\frac{θ}{1 - θ}$ at 95%.

I think this is related to the delta method where if X is a random variable then variance of g(X) is given as $σ^2$  times $g'(X)^2$.

If I'm right then $g'(X)^2$ = $\frac{1}{(1 - θ)^4}$.

Τhen we have $\frac{θ(1 - θ)}{T}$ as the estimator of the $σ^2$ and this is multiplied by $g'(X)^2$ = $\frac{1}{(1 - θ)^4}$

Then the asymptotic variance of is $\frac{θ(1 - θ)}{Τ}$ times $\frac{1}{(1 - θ)^4}$ = $\frac{θ}{Τ(1 - θ)^3}$

Is this correct?

If so then the 95% CI would be $\hat{θ} \pm 1.96 \sqrt{\frac{θ}{T(1 - θ)^3}}$

Or (again assuming the variance is correct) would the CI be

$\frac{\hat{θ}}{1 - \hat{θ}} \pm 1.96 \sqrt{\frac{θ}{T(1 - θ)^3}}$ ?

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  • $\begingroup$ When I first wrote it I used the Binomial distribution and the result had a T in the numerator. I thought perhaps it should be the Bernoulli distribution since we are looking at each $y_t$ as a Bernoulli. $\endgroup$ – MHall Oct 21 '18 at 0:07
  • $\begingroup$ After density, need to derive how to estimate $θ$. derived the variance of this estimate. Then consider the function of $θ$. $\endgroup$ – user158565 Oct 21 '18 at 0:40
  • $\begingroup$ I updated the question to include the derivations of the estimator of θ and the variance of that estimate. $\endgroup$ – MHall Oct 21 '18 at 2:55
  • $\begingroup$ I suggest studying the Wilson confidence interval for proportions. $\endgroup$ – Frank Harrell Oct 21 '18 at 3:12
  • $\begingroup$ I rewrite the second half in answer. But $g(θ)=\frac θ{1−θ}$ is monotonic function of $θ$, so get the CI for $θ$ first and convert CI by g function is good method. $\endgroup$ – user158565 Oct 21 '18 at 3:29
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By the delta method, if $X$ is a random variable with $\mathrm{Var}(X) = \sigma^2$ then $\mathrm{Var}(f(X)) = \sigma^2 [g′(X)]^2$.

For $g(θ) = \frac θ{1−θ}$, we can estimate $g(θ)$ by $\hat{g(θ)} = \frac {\hat θ}{1−\hat θ}$

Then by the delta method, $\mathrm{Var}(\hat{g(θ)}) = \mathrm{Var}(\hat θ)[g'(θ)]^2 =\frac{θ(1-θ)}T\frac 1{(1-θ)^4} = \frac θ {T(1-θ)^3} $

The 95% CI of $g(θ)$ can be get by $\frac {\hat θ}{1−\hat θ} \pm 1.96\sqrt{\frac {\hat {θ}} {T(1-\hat {θ})^3}}$

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  • $\begingroup$ .... 1.86? .... $\endgroup$ – Glen_b Oct 21 '18 at 7:08

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